Question

PLease Help!! Does the sequence sin(n) have a convergent subsequence? Please Justify

Answer 1

Every bounded sequence have a convergent sub-sequence...remember that

Answer 2

With that you can construct the proof.

Answer 3

But I'm not sure how to get the subsequence

Answer 4

well... thats a trivial case lol

Answer 5

I'm assuming he wants a non-trivial case. but yes that works too

Answer 6

So it's a rule that sin n = 2*pi*k

Answer 7

but how did you get n=2*pi*k ?

Answer 8

dude... look at what 2 Pi k is for sin

Answer 9

So therefore it oscillates between 1 and -1 as n approaches infinity ?

Answer 10

And isnt bounded therefore diverges ?

Answer 11

The Bolzano-Weierstrass Theorem: Every bounded sequence of real numbers has a
convergent subsequence.

Answer 12

I'm assuming n was real

Answer 13

Ohh... n isn't real... then pellet. lol

Answer 14

Well try this: Every sequence has a monotone sub-sequence, and if the sequence is bounded, then then monotone sub-sequence is also bounded and converges.

Answer 15

Alright I get it for this question ... But I'm still confused about the n=2*pi*k .. I dont know what was done with that or how it was gotten.. could you clarify on that please ?

Answer 16

Due to of course all monotonic bounded sequences converging.... that should be a sufficient proof.

Answer 17

But it doesnt have a sub sequence in this case correct ?

Answer 18

The sequence sin(n) is bounded, since -1 ≤ sin(n) ≤ 1 for any value of n.

The Bolzano-Weierstrass theorem tells you this must have a convergence subsequence.


In general, take any convergent sequence a[n] → L.

Then consider any random sequence, probably not convergent, b[n].



Then define the "zipper" of the two to be c[n], where:

c[2n] = a[n]
c[2n+1] = b[n]


Then c[2n] is a convergent subsequence of c[n], since c[2n] = a[n].

Notice how the b[n] could do any sort of whacky thing, and they would not affect this subsequence. So in general, different subsequences can converge to many different things, or diverge, or whatever else -- different kinds of subsequences can "coexist" in a sequence.


There are any cases in analysis when a sequence needs to just be well-behaved (e.g. bounded) and then a convergent subsequence must exist. These can be used in other proofs to derive stronger, better theorems.

Answer 19

Ok I get it .. Thanks guys!!! really appreciate the help !!